meta user
network profile
| bio | website | poisson.phc.unipi.it/~mossa |
|---|---|---|
| location | Earth | |
| age | 25 | |
| visits | member for | 1 year, 11 months |
| seen | 12 hours ago | |
| stats | profile views | 220 |
I'm math student, in particular I'm interested in algebra, geometry, topology and category theory (especially higher dimensional category theory) and its application in mathematics.
|
Jan 7 |
revised |
Foundation for category theory made some improvement to the question |
|
Jan 7 |
comment |
Foundation for category theory @IttayWeiss I've edited the question. |
|
Jan 7 |
revised |
Foundation for category theory made some improvement to the question |
|
Jan 7 |
comment |
Foundation for category theory @IttayWeiss good point....give me some time. I'll try to reformulate the question in a different way. Meanwhile thanks for the comments. |
|
Jan 7 |
comment |
Foundation for category theory @IttayWeiss Not exactly, I'm looking for concrete categories in which develop category theory. For what I got topos theory require category theory and set theory as its basis to be fully developed. What I'm asking for is something that we could call a foundational category. Hope I've been able to clear my doubts. |
|
Jan 7 |
answered | Looking for philosophical subject for my Bachelor Thesis |
|
Jan 7 |
asked | Foundation for category theory |
|
Jan 5 |
comment |
What can we say about the map $G\mapsto \text{Aut}(G)$ on the proper class of all groups? That's true, but how does it look like when we restrict it to a skeletal category? |
|
Jan 5 |
revised |
What can we say about the map $G\mapsto \text{Aut}(G)$ on the proper class of all groups? made a correction |
|
Jan 5 |
answered | What can we say about the map $G\mapsto \text{Aut}(G)$ on the proper class of all groups? |
|
Dec 30 |
comment |
When are elements in a tensor product equal to $0$? of course we suppose that $a,b$ as above must be not torsion element, i.e. both non null. :) |
|
Dec 30 |
comment |
When are elements in a tensor product equal to $0$? @FortuonPaendrag in your case $A \otimes_R B = \mathbb Z \otimes_\mathbb{Z} \mathbb Z = \mathbb Z$ but this doesn't seem trivial to me and more important for each pair of elements $a,b \in \mathbb Z$ you get that $a \otimes b = ab$ which is not $0$. |
|
Dec 30 |
comment |
On infinite groups that is not Simple @ChrisEagle You're right, but it seems that DonAntonio beat me in time. Anyway thank you both to helping me improving the answer. :) |
|
Dec 30 |
answered | On infinite groups that is not Simple |
|
Dec 22 |
asked | Lax algebras as lax morphisms |
|
Dec 19 |
awarded | Nice Answer |
|
Dec 10 |
comment |
Has this algebraic structure been named or studied? So basically what you're describing is a monoid with an endomorphism on it, right?! |
|
Dec 8 |
accepted | Reasons for coherence for bi/monoidal categories |
|
Dec 8 |
comment |
semidirect product, split extension @grendizer yes, that or more easily from the fact that $\beta \circ \gamma = \text{id}$, by splitting property. |
|
Dec 8 |
revised |
semidirect product, split extension improved answer |
